Originally Posted by
acfixerdude the graph of the original equation rotated around the x axis kinda looks like a football... if I take shells to calculate volume, then I have shells centered around x=3 and from -y to y in height. So the height is 2y and the width is going to be 2pi*r where r=3-x if we're differentiating from 2 to 3. substitute for y and we get the integral from 2 to 3 of 2(-x^2+6x-8)*2pi(3-x)dx = pi
using disks, the area of one disk is going to be pi*r^2 where r is going to be equal to y. A=pi(-x^2+6x-8)^2.... I set up the integral from 2 to 4 of pi(-x^2+6x-8)^2 dx = 16pi/15...
What am I doing wrong?